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To solve the problem of finding the velocity of the object at [tex]\( t = 2 \)[/tex] seconds given the acceleration function and initial velocity, we follow these steps:
1. Define the acceleration function [tex]\( a(t) \)[/tex]:
Given the acceleration function [tex]\( a(t) = (3t + 3)^{-4} \)[/tex].
2. Integrate the acceleration function to find the velocity function [tex]\( v(t) \)[/tex]:
To find the velocity function [tex]\( v(t) \)[/tex], we need to integrate the acceleration function with respect to time [tex]\( t \)[/tex]:
[tex]\[ v(t) = \int a(t) \, dt = \int (3t + 3)^{-4} \, dt \][/tex]
3. Perform the integration:
Integrate [tex]\( \int (3t + 3)^{-4} \, dt \)[/tex]. This can be done using integration techniques such as substitution.
Let [tex]\( u = 3t + 3 \)[/tex], hence [tex]\( du = 3 \, dt \)[/tex] or [tex]\( dt = \frac{du}{3} \)[/tex].
Substituting in, we get:
[tex]\[ \int (3t + 3)^{-4} \, dt = \int u^{-4} \, \frac{du}{3} = \frac{1}{3} \int u^{-4} \, du \][/tex]
The integral of [tex]\( u^{-4} \)[/tex] is:
[tex]\[ \int u^{-4} \, du = \frac{u^{-3}}{-3} = -\frac{1}{3} u^{-3} \][/tex]
Substituting back [tex]\( u = 3t + 3 \)[/tex]:
[tex]\[ -\frac{1}{3} \cdot \frac{1}{3(3t + 3)^3} = -\frac{1}{9(3t + 3)^3} \][/tex]
Therefore, the velocity function [tex]\( v(t) \)[/tex] is:
[tex]\[ v(t) = -\frac{1}{9(3t + 3)^3} + C \][/tex]
4. Determine the constant of integration [tex]\( C \)[/tex] using the initial condition:
Given that the velocity at [tex]\( t = 0 \)[/tex] is 4 meters per second:
[tex]\[ v(0) = 4 = -\frac{1}{9(3 \cdot 0 + 3)^3} + C = -\frac{1}{9 \cdot 27} + C = -\frac{1}{243} + C \][/tex]
Solving for [tex]\( C \)[/tex]:
[tex]\[ 4 = -\frac{1}{243} + C \implies C = 4 + \frac{1}{243} = \frac{972 + 1}{243} = \frac{973}{243} \][/tex]
Therefore, the function [tex]\( v(t) \)[/tex] becomes:
[tex]\[ v(t) = -\frac{1}{9(3t + 3)^3} + \frac{973}{243} \][/tex]
5. Calculate the velocity at [tex]\( t = 2 \)[/tex]:
Substitute [tex]\( t = 2 \)[/tex] into the velocity function:
[tex]\[ v(2) = -\frac{1}{9(3 \cdot 2 + 3)^3} + \frac{973}{243} = -\frac{1}{9 \cdot (6 + 3)^3} + \frac{973}{243} = -\frac{1}{9 \cdot 729} + \frac{973}{243} = -\frac{1}{6561} + \frac{973}{243} \][/tex]
Find a common denominator for these fractions:
[tex]\[ \frac{-1}{6561} + \frac{973 \cdot 27}{6561} = \frac{-1 + 26271}{6561} = \frac{26243}{6561} \][/tex]
6. Simplify and round to six decimal places:
Simplifying the fraction [tex]\(\frac{26243}{6561}\)[/tex] and converting to decimal form:
[tex]\[ v(2) \approx 3.999848 \][/tex]
Therefore, the velocity of the object 2 seconds later is approximately [tex]\(3.999848\)[/tex] meters per second.
1. Define the acceleration function [tex]\( a(t) \)[/tex]:
Given the acceleration function [tex]\( a(t) = (3t + 3)^{-4} \)[/tex].
2. Integrate the acceleration function to find the velocity function [tex]\( v(t) \)[/tex]:
To find the velocity function [tex]\( v(t) \)[/tex], we need to integrate the acceleration function with respect to time [tex]\( t \)[/tex]:
[tex]\[ v(t) = \int a(t) \, dt = \int (3t + 3)^{-4} \, dt \][/tex]
3. Perform the integration:
Integrate [tex]\( \int (3t + 3)^{-4} \, dt \)[/tex]. This can be done using integration techniques such as substitution.
Let [tex]\( u = 3t + 3 \)[/tex], hence [tex]\( du = 3 \, dt \)[/tex] or [tex]\( dt = \frac{du}{3} \)[/tex].
Substituting in, we get:
[tex]\[ \int (3t + 3)^{-4} \, dt = \int u^{-4} \, \frac{du}{3} = \frac{1}{3} \int u^{-4} \, du \][/tex]
The integral of [tex]\( u^{-4} \)[/tex] is:
[tex]\[ \int u^{-4} \, du = \frac{u^{-3}}{-3} = -\frac{1}{3} u^{-3} \][/tex]
Substituting back [tex]\( u = 3t + 3 \)[/tex]:
[tex]\[ -\frac{1}{3} \cdot \frac{1}{3(3t + 3)^3} = -\frac{1}{9(3t + 3)^3} \][/tex]
Therefore, the velocity function [tex]\( v(t) \)[/tex] is:
[tex]\[ v(t) = -\frac{1}{9(3t + 3)^3} + C \][/tex]
4. Determine the constant of integration [tex]\( C \)[/tex] using the initial condition:
Given that the velocity at [tex]\( t = 0 \)[/tex] is 4 meters per second:
[tex]\[ v(0) = 4 = -\frac{1}{9(3 \cdot 0 + 3)^3} + C = -\frac{1}{9 \cdot 27} + C = -\frac{1}{243} + C \][/tex]
Solving for [tex]\( C \)[/tex]:
[tex]\[ 4 = -\frac{1}{243} + C \implies C = 4 + \frac{1}{243} = \frac{972 + 1}{243} = \frac{973}{243} \][/tex]
Therefore, the function [tex]\( v(t) \)[/tex] becomes:
[tex]\[ v(t) = -\frac{1}{9(3t + 3)^3} + \frac{973}{243} \][/tex]
5. Calculate the velocity at [tex]\( t = 2 \)[/tex]:
Substitute [tex]\( t = 2 \)[/tex] into the velocity function:
[tex]\[ v(2) = -\frac{1}{9(3 \cdot 2 + 3)^3} + \frac{973}{243} = -\frac{1}{9 \cdot (6 + 3)^3} + \frac{973}{243} = -\frac{1}{9 \cdot 729} + \frac{973}{243} = -\frac{1}{6561} + \frac{973}{243} \][/tex]
Find a common denominator for these fractions:
[tex]\[ \frac{-1}{6561} + \frac{973 \cdot 27}{6561} = \frac{-1 + 26271}{6561} = \frac{26243}{6561} \][/tex]
6. Simplify and round to six decimal places:
Simplifying the fraction [tex]\(\frac{26243}{6561}\)[/tex] and converting to decimal form:
[tex]\[ v(2) \approx 3.999848 \][/tex]
Therefore, the velocity of the object 2 seconds later is approximately [tex]\(3.999848\)[/tex] meters per second.
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